What is the extraneous solution to these equations? $\dfrac{x^2 + 12}{x - 10} = \dfrac{8x + 32}{x - 10}$
Multiply both sides by $x - 10$ $ \dfrac{x^2 + 12}{x - 10} (x - 10) = \dfrac{8x + 32}{x - 10} (x - 10)$ $ x^2 + 12 = 8x + 32$ Subtract $8x + 32$ from both sides: $ x^2 + 12 - (8x + 32) = 8x + 32 - (8x + 32)$ $ x^2 + 12 - 8x - 32 = 0$ $ x^2 - 20 - 8x = 0$ Factor the expression: $ (x - 10)(x + 2) = 0$ Therefore $x = 10$ or $x = -2$ At $x = 10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 10$, it is an extraneous solution.